# Lecture 7 – The giant component

I am aiming to write a short post about each lecture in my ongoing course on Random Graphs. Details and logistics for the course can be found here.

As we edge into the second half of the course, we are now in a position to return to the question of the phase transition between the subcritical regime $\lambda<1$ and the supercritical regime $\lambda>1$ concerning the size of the largest component $L_1(G(n,\lambda/n))$.

In Lecture 3, we used the exploration process to give upper bounds on the size of this largest component in the subcritical regime. In particular, we showed that

$\frac{1}{n}\big| L_1(G(n,\lambda/n)) \big| \stackrel{\mathbb{P}}\rightarrow 0.$

If we used slightly stronger random walk concentration estimates (Chernoff bounds rather than 2nd-moment bounds from Chebyshev’s inequality), we could in fact have shown that with high probability the size of this largest component was at most some logarithmic function of n.

In this lecture, we turn to the supercritical regime. In the previous lecture, we defined various forms of weak local limit, and asserted (without attempting the notationally-involved combinatorial calculation) that the random graph $G(n,\lambda/n)$ converges locally weakly in probability to the Galton-Watson tree with $\text{Poisson}(\lambda)$ offspring distribution, as we’ve used informally earlier in the course.

Of course, when $\lambda>1$, this branching process has strictly positive survival probability $\zeta_\lambda>0$. At a heuristic level, we imagine that all vertices whose local neighbourhood is ‘infinite’ are in fact part of the same giant component, which should occupy $(\zeta_\lambda+o_{\mathbb{P}}(1))n$ vertices. In its most basic form, the result is

$\frac{1}{n}\big|L_1(G(n,\lambda/n))\big|\;\stackrel{\mathbb{P}}\longrightarrow\; \zeta_\lambda,\quad \frac{1}{n}\big|L_2(G(n,\lambda/n))\big| \;\stackrel{\mathbb{P}}\longrightarrow\; 0,$ (*)

where the second part is a uniqueness result for the giant component.

The usual heuristic for proving this result is that all ‘large’ components must in fact be joined. For example, if there are two giant components, with sizes $\approx \alpha n,\approx \beta n$, then each time we add a new edge (such an argument is often called ‘sprinkling‘), the probability that these two components are joined is $\approx 2ab$, and so if we add lots of edges (which happens as we move from edge probability $\lambda-\epsilon\mapsto \lambda$ ) then with high probability these two components get joined.

It is hard to make this argument rigorous, and the normal approach is to show that with high probability there are no components with sizes within a certain intermediate range (say between $\Theta(\log n)$ and $n^\alpha$) and then show that all larger components are the same by a joint exploration process or a technical sprinkling argument. Cf the books of Bollobas and of Janson, Luczak, Rucinski. See also this blog post (and the next page) for a readable online version of this argument.

I can’t find any version of the following argument, which takes the weak local convergence as an assumption, in the literature, but seems appropriate to this course. It is worth noting that, as we shall see, the method is not hugely robust to adjustments in case one is, for example, seeking stronger estimates on the giant component (eg a CLT).

Anyway, we proceed in three steps:

Step 1: First we show, using the local limit, that for any $\epsilon>0$,

$\frac{1}{n}\big|L_1(G(n,\lambda/n))\big| \le \zeta_\lambda+\epsilon,$ with high probability as $n\rightarrow\infty$.

Step 2: Using a lower bound on the exploration process, for $\epsilon>0$ small enough

$\frac{1}{n}\big|L_1(G(n,\lambda/n))\big| \ge \epsilon,$ with high probability.

Step 3: Motivated by duality, we count isolated vertices to show

$\mathbb{P}(\epsilon n\le |L_1| \le (\zeta_\lambda-\epsilon)n) \rightarrow 0.$

We will return to uniqueness at the end.

Step 1

This step is unsurprising. The local limit gives control on how many vertices are in small components of various sizes, and so gives control on how many vertices are in small components of all finite sizes (taking limits in the right order). This gives a bound on how many vertices can be in the giant component.

(Note: parts of this argument appear in the text and exercises of Section 1.4 in the draft of Volume II of van der Hofstad’s notes, which can be found here.)

We can proceed in greater generality, by considering a sequence of random graphs $G_n$ which converge locally weakly in probability to T, a random tree, with survival probability $\zeta=\mathbb{P}(|T|=\infty)>0$. We will show that:

Proposition: $\mathbb{P}(L_1(G_n)\ge (\zeta+\epsilon)n) \rightarrow 0,$ for each $\epsilon>0$.

As a preliminary, note that for every $k\in\mathbb{N}$, there are finitely many rooted graphs $(H,\rho_H)$ with size k. We can also identify whether a graph has size k by looking at a ball of radius r>k around any vertex. In particular, by summing over all graphs with size k, the weak local limit implies:

$\frac{1}{n}\sum_{v\in[n]} \mathbf{1}_{\{|C^{G_n}(\rho_n)|=k\}} = \frac{1}{n} \sum_{|V(H)|=k} \sum_{v\in[n]} \mathbf{1}_{\{B_r^{G_n}(\rho_n)\simeq (H,\rho_H)\}}$

$\stackrel{\mathbb{P}}\longrightarrow \;\sum_{|V(H)|=k} \mathbb{P}(B_r^T(\rho)\simeq (H,\rho_H)) = \mathbb{P}(|T|=k).$

Furthermore, we can then control the tail as

$\frac{1}{n}\sum_{v\in[n]} \mathbf{1}_{\{|C^{G_n}(v)|\ge k\}}\;\stackrel{\mathbb{P}}\longrightarrow \mathbb{P}(|T|\ge k).$

(Recall that the LHS of this statement is the proportion of vertices in components of size at least k.)

We will make the trivial but useful observation that in any graph the largest component has size at least k precisely if at least k vertices are in components of size at least k (!). Ie

$|L_1(G)|\ge k\quad\iff\quad \sum_{v\in[n]} \mathbf{1}{\{C^G(v)|\ge k\}} \ge k.$

Returning now to the problem at hand, we have $\mathbb{P}(|T|\ge k)\downarrow\zeta$ as $k\rightarrow\infty$, so we may pick k such that $\mathbb{P}(|T|\ge k)<\zeta+\epsilon$.

But then, using our ‘trivial but useful’ observation:

$\mathbb{P}(L_1(G_n)\ge (\zeta+\epsilon)n) = \mathbb{P}(\sum_{v\in[n]} \mathbf{1}_{\{|C^{G_n}(v)|\ge (\zeta+\epsilon)n \}} \ge (\zeta+\epsilon)n)$

$\le \mathbb{P}(\frac{1}{n}\sum_{v\in[n]} \mathbf{1}_{\{|C^{G_n}(v)|\ge k\}} \ge \zeta+\epsilon).$ (**)

Note that we have replaced $(\zeta+\epsilon)n$ by k in this final step for a bound. However, the random quantity inside the probability is known to converge in probability to $\mathbb{P}(|T|\ge k)<\zeta+\epsilon$. So in fact this probability (**) vanishes as $n\rightarrow\infty$.

Step 2

Remember the exploration process, where $v=v_1,v_2,\ldots,v_n$ is a labelling of the vertices of $G(n,\lambda/n)$ in breadth-first order. Defining

$X_i:= \#\{w\in[n]\,:\, w\in\Gamma(v_i),\,w\not\in \Gamma(v_j),\,j\in[i-1]\},$

the number of children of vertex $v_i$, we set

$S_0:=0,\quad S_i:=S_{i-1}+(X_i-1),\; i\ge 1,$

to be (a version of) the exploration process. It will be useful to study

$H_0:=0,\quad H_k:=\min\{i\,:\, S_i=-k\},$

the hitting times of (-k), as then $\{v_{H_{k-1}+1},\ldots,v_{H_k}\}$ is the kth component to be explored.

Unlike for a tree, we have multiple components, and essentially the process decreases by one each time we start a new component, which means that the current value no longer describes the number of vertices on the stack. In general, this is given by $S_i - \min_{0\le j\le i}S_j$, and so

$X_i\stackrel{d}= \text{Bin}(n-i-(S_{i-1}-\min_{0\le j\le i-1}S_j),\, \frac{\lambda}{n}),$

which we may stochastically bound below by

$\ge_{st} \text{Bin}(n-2i-S_{i-1},\,\frac{\lambda}{n}),$

noting that this is extremely crude.

We want to study whether $S_i$ ever exceeds $\epsilon n$, for some $\epsilon>0$ to be determined later.

For reasons that will become clear in the following deduction, it’s convenient to fix $\alpha>0$ small such that $\lambda(1-2\alpha)>1$, and then choose $\epsilon>0$ such that

$\alpha\left[\lambda(1-2\alpha-\epsilon)-1\right]>\epsilon.$

(which is possible by continuity since the given relation holds when $\epsilon=0$.)

Now, when $i\le \alpha n$ and $S_{i-1}\le \epsilon n$, we have

$X_i\ge_{st} \text{Bin}(n(1-2\alpha-\epsilon),\frac{\lambda}{n}).$

The following argument requires some kind of submartingale approach (involving coupling with a simpler process at the stopping time) to make rigorous, which is beyond the scope of this course’s prerequisites.

However, informally, if we assume that $\max_{i\le \alpha n} S_i\le \epsilon n$, ‘then’

$\frac{1}{n}S_{\alpha n}\ge_{st} \frac{1}{n}\text{Bin}(\alpha n\cdot n(1-2\alpha-\epsilon),\,\frac{\lambda}{n}) - \alpha.$

But this distribution is concentrated on a value which is, by our obscure assumption, $>\epsilon$ (!) contradicting the assumption on the maximum. Thus we conclude that

$\mathbb{P}(\max_{i\le \alpha n} S_i\le \epsilon n)\rightarrow 0,$ as $n\rightarrow\infty.$

We conclude that $\max_{i\le \alpha n} S_i\ge\epsilon n$ holds with high probability. But remember that $S_{i+1}\ge S_i -1$ so if $S_i\ge \epsilon n$, then all of $S_i,S_{i+1},\ldots,S_{i+\lfloor \epsilon n\rfloor}$ are non-negative, and so certainly $v_i,v_{i+1},\ldots,v_{i+\lfloor \epsilon n\rfloor}$ are in the same component of the graph, and $L_1(G(n,\lambda/n))\ge \epsilon n$ with high probability.

Step 3

The motivation for this section is duality. Recall (from Lecture 5) that if we condition a supercritical Poisson GW tree on extinction, we obtain the distribution of a dual subcritical Poisson GW tree. This relation moves across to the world of the sparse Erdos-Renyi random graph. If you exclude the giant component, you are left with a subcritical random graph (on a smaller vertex set), and this applies equally well to the local limits. Essentially, if we exclude a component, and take the local limit of what remains, we get the wrong answer unless the component we excluded was a giant component with size $\approx \zeta_\lambda n$, or was small.

As we shall see, this effect is captured sufficiently by counting isolated vertices.

First, we state a Fact: when $1-\zeta_\lambda , then $ye^{-\lambda y}>e^{-\lambda}$. This convexity property is easily checked by comparing derivatives, and will be useful shortly.

Now, we study $I_n$, the number of isolated vertices in $G_n$, under conditioning that $\{1,2,\ldots,k\}$ is a component for various values k. Note that unless k=1, we have

$\mathbb{E}[I_n\,\big|\, \{1,2,\ldots,k\}\text{ a cpt}] = (n-k)(1-\frac{\lambda}{n})^{n-k-1},$

for exactly the same reason as when we did this calculation for the original graph several lectures back. We will consider k in the range $\epsilon n \le k\le (\zeta_\lambda - \epsilon) n$.

We can take a limit of this expectation in appropriate uniformly using the Fact above, since the function $ye^{-\lambda y}$ is suitably well-behaved, to obtain

$\liminf_{n\rightarrow\infty} \frac{1}{n}\min_{\epsilon n\le k\le (\zeta-\epsilon)n} (n-k)(1-\frac{\lambda}{n})^{n-k-1}$

$\ge \min_{\epsilon \le x \le \zeta-\epsilon} (1-x)e^{-\lambda(1-x)}\ge e^{-\lambda}+\epsilon',$

where $\epsilon'>0$. So

$\liminf_{n\rightarrow\infty} \min_{\epsilon n\le k\le (zeta-\epsilon)n} \frac{1}{n}\mathbb{E}\left[ I_n\,\big|\, \{1,\ldots,k\}\text{ a cpt}\right] \ge e^{-\lambda}+\epsilon'.$

But $\frac{I_n}{n}$ is bounded above (by 1, of course), and so lower bounds on the expectation give lower bounds on upper tail, leading to

$\liminf_{n\rightarrow\infty}\min_{\epsilon n\le k\le (\zeta-\epsilon)n} \mathbb{P}\left( \frac{I_n}{n}\ge e^{-\lambda}+\frac{\epsilon'}{2}\,\big|\, \{1,\ldots,k\}\text{ a cpt} \right) >0.$

However, we know $\frac{I_n}{n}\stackrel{\mathbb{P}}\rightarrow e^{-\lambda}$ (for example by local convergence…). Therefore, in order to make the unconditional probability vanish, the probability of the conditioning event in question must also vanish, ie

$\mathbb{P}\left(\epsilon n\le |C^{G_n}(1)|\le (\zeta_\lambda-\epsilon)n\right)\rightarrow 0.$

Finally, since

$\mathbb{P}\left(\epsilon n \le |C^{G_n}(1)|\le (\zeta_\lambda-\epsilon)n\right) \ge \frac{1}{\epsilon}\mathbb{P}(\epsilon n\le |L_1(G)|\le (\zeta_\lambda-\epsilon)n),$

the corresponding result holds for the largest component, not just the observed component.

Uniqueness and overall comments

Uniqueness can be obtained by a slight adjustment of Step 1. Morally, Step 1 is saying that a proportion asymptotically at most $\zeta_\lambda$ of the vertices are in large components, so it is possible (and an exercise in the course) to adjust the argument to show

$\frac{1}{n}|L_1|+\frac{1}{n}|L_2| \le \zeta_\lambda+\epsilon,$ with high probability,

from which the uniqueness result follows immediately.

In particular, it’s worth noting that this is an example of a bootstrapping argument, where we show a weak version of our goal result (in Step 2), but then use this to show the full result.

Note also that we can use the duality principle to show logarithmic bounds on the size of the second-largest component in exactly the same way that we showed logarithmic bounds on the size of the largest component in the subcritical regime. The whole point of duality is that these are the same problem!

# Lecture 4 – Hitting time theorem

I am aiming to write a short post about each lecture in my ongoing course on Random Graphs. Details and logistics for the course can be found here.

This lecture consisted of revision of the most relevant theory of Galton-Watson trees, with a focus on the case where the offspring distribution is Poisson, since, as we have seen in previous lectures, this is a strong candidate to approximate the structure of $G(n,\lambda/n)$. It makes sense to cover the theory of the trees before attempting to make rigorous the sense of approximation.

Given a Galton-Watson tree T, it is natural to label the vertices in a breadth-first order as $\varnothing=v_1,v_2,\ldots,v_{|T|}$. This is easiest if we have constructed the Galton-Watson tree as a subset of the infinite Ulam-Harris tree, where vertices have labels like (3,5,17,4), whose parent is (3,5,17). If this child vertex is part of the tree, then so are (3,5,17,1), (3,5,17,2), and (3,5,17,3). This means our breadth-first order is canonically well-defined, as we have a natural ordering of the children of each parent vertex.

Note: one advantage of using breadth-first order rather than depth-first order (which corresponds to the usual dictionary, or lexicographic ordering of the labels) is that if the tree is infinite, we don’t explore all of it during a depth-first search. (In the sense that there exist vertices which are never given a finite label.) For breadth-first search, a similar problem arises precisely when some vertex has infinitely many children. For a conventional Galton-Watson tree, the latter situation is much less of a problem than the infinite total population problem, which happens with positive probability whenever $\mu=\mathbb{E}[X]>1$.

Anyway, given the depth-first order, one can consider an exploration process $S_0,S_1,S_2,\ldots,S_{|T|}$ given by

$S_0=1,\quad S_i=S_{i-1}+(X_i-1),$

where $X_i$ is the number of children of $v_i$. In this way, we see that

$S_i=\big| \Gamma(v_1)\cup\ldots\cup\Gamma(v_i)\backslash \{v_1,\ldots,v_i\}\big|,\quad i\ge 1,$

records the number of vertices in some stack containing those which we have ‘seen but not explored’. Some authors prefer to start from 0, in which case one ends up with a similar but slightly different interpretation of the ‘stack’, but that’s fine since we aren’t going to define formally what ‘seen’ and ‘explored’ means in this post.

Essentially, we exhaust the vertices of the tree whenever $S_t=0$, and so the condition that $|T|=n$ requires

$S_n=0,\quad S_m\ge 1,\; m=0,1,\ldots,n-1.$

Conveniently, so long as we have avoiding ordering ambiguity, for example by insisting that trees live within the Ulam-Harris tree, we can reconstruct T uniquely from $(S_0,S_1,\ldots,S_{|T|})$.

Furthermore, if T is a Galton-Watson process, then the numbers of children $X_i$ are IID, and so in fact this exploration process is a random walk, and the size of the tree can be recovered as the hitting time of zero.

Note: making fully rigorous the argument that children in the GW tree are independent of the breadth-first walk fully rigorous is somewhat technical, and not to be dismissed lightly, though not of principle interest at the level of this topics course. See Proposition 1.5 in Section 1.2 of Le Gall’s notes or Section 1.2.2 of my doctoral thesis for further discussion and argument.

The hitting time theorem allows us to study the distribution of the hitting time of a random walk whose increments are bounded below by -1, in terms of the distribution of the value of the random walk.

Theorem: Let $(S_n,\, n\ge 0)$ be a random walk with $S_0=0$ and IID increments $(X_n,n\ge 1)$ satisfying $\mathbb{P}(X_n\ge -1)=1$. Let $H_{-k}=\inf \left\{n\,:\, S_n=-k\right\}$ be the hitting time of -k.

Then $\mathbb{P}\big( H_{-k}=n\big) = \frac{k}{n}\mathbb{P}\big(S_n=-k)$.

Commentary: there are local central limit theorem estimates and large deviation estimates that allow good control of the probability on the RHS for a rich class of contexts. So at a meta-level, the hitting time theorem allows us to reduce a complicated (though still classical) problem, to a real classical problem, which is particularly helpful when the LHS is a device for capturing relevant information about our random tree model.

Proof context/history: The case k=1, as relevant to the trees, is often referred to as Dwass’s theorem, though the combinatorial argument pre-dates this, sometimes in the form of the similar ballot theoremA complete reference list would be challenging, but it certainly appears in Feller, due to Spitzer. In his recent book, van der Hofstad uses an argument by induction on n, noting that the statement is obvious when n=0,1,…,k-1, and obvious for different reasons when n=k. This argument can also be found published here.

I’m going to give a combinatorial argument, similar to one I’ve given in a post from years ago (back then, for the case k=1), because there wasn’t time to include this in the lecture.

Proof: from the point of view of the course, the most important probabilistic take home message from this argument is that we get significant control by introducing the randomness in two stages, and conditioning on the first round of randomness.

Here, what we do is to condition on the increments $(X_1,X_2,\ldots,X_n)_{\mathrm{cyc}}$ in cyclic order, ie without distinguishing between $(x_1,x_2,\ldots,x_n)$ and $(x_4,x_5,\ldots,x_n,x_1,x_2,x_3)$. We’ll declare which of these is the true ordering at the end. Anyway, the punchline is the following lemma.

Lemma: when $x_1+\ldots +x_n=-k$, and all $x_i\ge -1$, then exactly k of the n orderings corresponding to the cyclic ordering $(x_1,\ldots,x_n)_{\mathrm{cyc}}$ have the property that $x_1+\ldots+x_\ell \ge -k+1$.

With this lemma, it’s clear why the hitting time theorem holds. We condition on $(x_1,\ldots,x_n)_{\mathrm{cyc}}$. If the sum of these elements is not -k, then there’s no chance that the hitting time is n. If the sum is -k, then when we choose a true ordering at random, we get the hitting time as n with conditional probability k/n.

Proof of lemma: At all times, take indices modulo n. We want labels $\alpha$ satisfying:

$x_\alpha,\; x_\alpha+x_{\alpha+1},\; \ldots\;, ;\ x_{\alpha}+\ldots+x_{\alpha+n-2}\text{ are all }\ge -(k-1).$ (*)

Step 1: Suppose there are $\alpha_1<\alpha_2<\ldots<\alpha_{k+1}$ satisfying (*). It’s helpful to think about going round the circle of values. If we go all the way round the circle, we get -k, but if we go from $x_{\alpha_{i+1}}\to x_{\alpha_i-1}$ then we have an initial segment from $x_{\alpha_{i+1}}$, and so

$x_{\alpha_i}+\ldots+x_{\alpha_{i+1}-1} = -k - (x_{\alpha_{i+1}}+\ldots+x_{\alpha_i-1}) \le -k - \left(-(k-1)\right) = -1.$

However, if we repeat this for the k+1 such blocks, we end up going all the way round the circle, and get

$\left(x_{\alpha_1}+\ldots+x_{\alpha_2-1}\right) + \left(x_{\alpha_2}+\ldots+x_{\alpha_3-1}\right) + \ldots$

$\ldots+ \left(x_{\alpha_{k+1}} + \ldots+x_{\alpha_1-1}\right) \le -(k+1),$

which is a contradiction since the total sum is k. So we have at most k such values of $\alpha$.

Step 2: Now we show that there is at least one such $\alpha$. Let

$\beta=\mathrm{argmin}_{1\le m\le n} (x_1+\ldots+x_m),$

and if there are multiple such m, let $\beta$ be the minimal such. Now set $\alpha=\beta+1$, and we will show this satisfies the requirements.

• Then $x_\alpha+x_{\alpha+1}+\ldots+x_{\alpha+m}\ge 0$, whenever $\alpha+m\le n$.
• And when $m<\beta$, we have

$x_{\alpha}+\ldots+x_n+x_1+\ldots+x_m = -k+(x_1+\ldots+x_m)-(x_1+\ldots+x_\beta).$

And by assumption, the first bracket is strictly greater than the second bracket, so

$x_\alpha + \ldots+x_n+x_1+\ldots+x_m \ge -(k-1).$

Step 3: Now we have an $\alpha$ which satisfies, we may assume WLOG that it is $\alpha_0=1$. Now we construct the hitting times:

$\alpha_1=\min\big(m\,:\, x_1+\ldots+x_m=-1\big),$

$\alpha_{k-1}=\min\big(m\,:\, x_1+\ldots+x_m=-(k-1)\big).$

We claim that all these also satisfy the requirements for $\alpha$. This is fairly clear from a diagram, or from the fact that when $\alpha_i\le m<\alpha_{i+1}$, then

$x_{\alpha_k}+x_{\alpha_k+1}+\ldots+x_m\ge x_{\alpha_k}+\ldots+x_{\alpha_i}= i-k,$

(or $i-k+n if i).

# BMO1 2018

The first round of the British Mathematical Olympiad was sat yesterday. The paper can be found here, and video solutions here. Copyright for the questions is held by BMOS. They are reproduced here with permission.

I hope any students who sat the paper enjoyed at least some of the questions, and found it challenging! The following commentaries on the problems are not official solutions, and are not always full solutions at all, but contain significant steps of solutions, so would be best saved until after you have attempted the problems, if you are planning to do so. I’ve written quite a lot about Q5 because I found it hard (or at least time-consuming) and somewhat atypical, and I’ve written a lot about Q6 because there was a lot to say. I hope at least some of this is interesting to some readers of all levels of olympiad experience.

Question 1

A list of five two-digit positive integers is written in increasing order on a blackboard. Each of the five integers is a multiple of 3, and each digit {0,1,…,9} appears exactly once on the blackboard. In how many ways can this be done? (Note that a two-digit number cannot begin with zero.)

It’s a trope of BMO1 that the first question must be doable by some sort of exhaustive calculation or listing exercise. Of course, that is rarely the most efficient solution.

However, there is normally a trade-off between eliminating all listing, and reducing to a manageable task.

The key observation here is that writing the integers in increasing order is really just a way to indicate that order of the choices doesn’t matter. Even if that seems counter-intuitive. The question wants to know how many ways to choose these five numbers. The order of choice doesn’t matter since we’re going to put them in ascending order on the blackboard anyway.

You want to make your choices with as much independence as possible. So it would, for example, be a bad idea to choose the smallest number first. How many possibilities are there where the smallest number is 24? What about 42? What about 69? These are all different, and some are zero, so will make the computation very taxing.

However, you might notice that the digits {0,3,6,9} have to go together to form two numbers, and the rest have to pair up with one digit from {1,4,7} and one from {2,5,8}. You might know that an integer is divisible by 3 precisely if its digit sum is divisible by 3, but in this context you wouldn’t lose too much time by simply listing everything! These tasks are now completely separate, so you can take the number of ways to pair up {0,3,6,9} and multiply by the number of ways to pair up {1,4,7} and {2,5,8}. You need to take care over the ordering. It does (obviously) matter which is the first digit and which is the second digit in a number!

# Random walks conditioned to stay positive

In this post, I’m going to discuss some of the literature concerning the question of conditioning a simple random walk to lie above a line with fixed gradient. A special case of this situation is conditioning to stay non-negative. Some notation first. Let $(S_n)_{n\ge 0}$ be a random walk with IID increments, with distribution X. Take $\mu$ to be the expectation of these increments, and we’ll assume that the variance $\sigma^2$ is finite, though at times we may need to enforce slightly stronger regularity conditions.

(Although simple symmetric random walk is a good example for asymptotic heuristics, in general we also assume that if the increments are discrete they don’t have parity-based support, or any other arithmetic property that prevents local limit theorems holding.)

We will investigate the probability that $S_n\ge 0$ for n=0,1,…,N, particularly for large N. For ease of notation we write $T=\inf\{n\ge 0\,:\, S_n<0\}$ for the hitting time of the negative half-plane. Thus we are interested in $S_n$ conditioned on T>N, or T=N, mindful that these might not be the same. We will also discuss briefly to what extent we can condition on $T=\infty$.

In the first paragraph, I said that this is a special case of conditioning SRW to lie above a line with fixed gradient. Fortunately, all the content of the general case is contained in the special case. We can repose the question of $S_n$ conditioned to stay above $n\alpha$ until step N by the question of $S_n-n\alpha$ (which, naturally, has drift $\mu-\alpha$) conditioned to stay non-negative until step N, by a direct coupling.

Applications

Simple random walk is a perfectly interesting object to study in its own right, and this is a perfectly natural question to ask about it. But lots of probabilistic models can be studied via naturally embedded SRWs, and it’s worth pointing out a couple of applications to other probabilistic settings (one of which is the reason I was investigating this literature).

In many circumstances, we can desribe random trees and random graphs by an embedded random walk, such as an exploration process, as described in several posts during my PhD, such as here and here. The exploration process of a Galton-Watson branching tree is a particularly good example, since the exploration process really is simple random walk, unlike in, for example, the Erdos-Renyi random graph G(N,p), where the increments are only approximately IID. In this setting, the increments are given by the offspring distribution minus one, and the hitting time of -1 is the total population size of the branching process. So if the expectation of the offspring distribution is at most 1, then the event that the size of the tree is large is an atypical event, corresponding to delayed extinction. Whereas if the expectation is greater than one, then it is an event with limiting positive probability. Indeed, with positive probability the exploration process never hits -1, corresponding to survival of the branching tree. There are plenty of interesting questions about the structure of a branching process tree conditional on having atypically large size, including the spine decomposition of Kesten [KS], but the methods described in this post can be used to quantify the probability, or at least the scale of the probability of this atypical event.

In my current research, I’m studying a random walk embedded in a construction of the infinite-volume DGFF pinned at zero, as introduced by Biskup and Louidor [BL]. The random walk controls the gross behaviour of the field on annuli with dyadically-growing radii. Anyway, in this setting the random walk has Gaussian increments. (In fact, there is a complication because the increments aren’t exactly IID, but that’s definitely not a problem at this level of exposition.) The overall field is decomposed as a sum of the random walk, plus independent DGFFs with Dirichlet boundary conditions on each of the annuli, plus asymptotically negligible corrections from a ‘binding field’. Conditioning that this pinned field be non-negative up to the Kth annulus corresponds to conditioning the random walk to stay above the magnitude of the minimum of each successive annular DGFF. (These minima are random, but tightly concentrated around their expectations.)

Conditioning on $\{T > N\}$

When we condition on $\{T>N\}$, obviously the resulting distribution (of the process) is a mixture of the distributions we obtain by conditioning on each of $\{T=N+1\}, \{T=N+2\},\ldots$. Shortly, we’ll condition on $\{T=N\}$ itself, but first it’s worth establishing how to relate the two options. That is, conditional on $\{T>N\}$, what is the distribution of T?

Firstly, when $\mu>0$, this event always has positive probability, since $\mathbb{P}(T=\infty)>0$. So as $N\rightarrow\infty$, the distribution of the process conditional on $\{T>N\}$ converges to the distribution of the process conditional on survival. So we’ll ignore this for now.

In the case $\mu\le 0$, everything is encapsulated in the tail of the probabilities $\mathbb{P}(T=N)$, and these tails are qualitatively different in the cases $\mu=0$ and $\mu<0$.

When $\mu=0$, then $\mathbb{P}(T=N)$ decays polynomially in N. In the special case where $S_n$ is simple symmetric random walk (and N has the correct parity), we can check this just by an application of Stirling’s formula to count paths with this property. By contrast, when $\mu<0$, even demanding $S_N=-1$ is a large deviations event in the sense of Cramer’s theorem, and so the probability decays exponentially with N. Mogulskii’s theorem gives a large deviation principle for random walks to lie above a line defined on the scale N. The crucial fact here is that the probabilistic cost of staying positive until N has the same exponent as the probabilistic cost of being positive at N. Heuristically, we think of spreading the non-expected behaviour of the increments uniformly through the process, at only polynomial cost once we’ve specified the multiset of values taken by the increments. So, when $\mu<0$, we have

$\mathbb{P}(T\ge(1+\epsilon)N) \ll \mathbb{P}(T= N).$

Therefore, conditioning on $\{T\ge N\}$ in fact concentrates T on N+o(N). Whereas by contrast, when $\mu=0$, conditioning on $\{T\ge N\}$ gives a nontrivial limit in distribution for T/N, supported on $[1,\infty)$.

A related problem is the value taken by $S_N$, conditional on {T>N}. It’s a related problem because the event {T>N} depends only on the process up to time N, and so given the value of $S_N$, even with the conditioning, after time N, the process is just an unconditioned RW. This is a classic application of the Markov property, beloved in several guises by undergraduate probability exam designers.

Anyway, Iglehart [Ig2] shows an invariance principle for $S_N | T>N$ when $\mu<0$, without scaling. That is $S_N=\Theta(1)$, though the limiting distribution depends on the increment distribution in a sense that is best described through Laplace transforms. If we start a RW with negative drift from height O(1), then it hits zero in time O(1), so in fact this shows that conditonal on $\{T\ge N\}$, we have T= N +O(1) with high probability. When $\mu=0$, we have fluctuations on a scale $\sqrt{N}$, as shown earlier by Iglehart [Ig1]. Again, thinking about the central limit theorem, this fits the asymptotic description of T conditioned on T>N.

Conditioning on $T=N$

In the case $\mu=0$, conditioning on T=N gives

$\left[\frac{1}{\sqrt{N}}S(\lfloor Nt\rfloor ) ,t\in[0,1] \right] \Rightarrow W^+(t),$ (*)

where $W^+$ is a standard Brownian excursion on [0,1]. This is shown roughly simultaneously in [Ka] and [DIM]. This is similar to Donsker’s theorem for the unconditioned random walk, which converges after rescaling to Brownian motion in this sense, or Brownian bridge if you condition on $S_N=0$. Skorohod’s proof for Brownian bridge [Sk] approximates the event $\{S_N=0\}$ by $\{S_N\in[-\epsilon \sqrt{N},+\epsilon \sqrt{N}]\}$, since the probability of this event is bounded away from zero. Similarly, but with more technicalities, a proof of convergence conditional on T=N can approximate by $\{S_m\ge 0, m\in[\delta N,(1-\delta)N], S_N\in [-\epsilon \sqrt{N},+\epsilon\sqrt{N}]\}$. The technicalities here emerge since T, the first return time to zero, is not continuous as a function of continuous functions. (Imagine a sequence of processes $f^N$ for which $f^N(x)\ge 0$ on [0,1] and $f^N(\frac12)=\frac{1}{N}$.)

Once you condition on $T=N$, the mean $\mu$ doesn’t really matter for this scaling limit. That is, so long as variance is finite, for any $\mu\in\mathbb{R}$, the same result (*) holds, although a different proof is in general necessary. See [BD] and references for details. However, this is particularly clear in the case where the increments are Gaussian. In this setting, we don’t actually need to take a scaling limit. The distribution of Gaussian *random walk bridge* doesn’t depend on the mean of the increments. This is related to the fact that a linear transformation of a Gaussian is Gaussian, and can be seen by examining the joint density function directly.

Conditioning on $T=\infty$

When $\mu>0$, the event $\{T=\infty\}$ occurs with positive probability, so it is well-defined to condition on it. When $\mu\le 0$, this is not the case, and so we have to be more careful.

First, an observation. Just for clarity, let’s take $\mu<0$, and condition on $\{T>N\}$, and look at the distribution of $S_{\epsilon N}$, where $\epsilon>0$ is small. This is approximately given by

$\frac{S_{\epsilon N}}{\sqrt{N}}\stackrel{d}{\approx}W^+(\epsilon).$

Now take $\epsilon\rightarrow\infty$ and consider the RHS. If instead of the Brownian excursion $W^+$, we instead had Brownian motion, we could specify the distribution exactly. But in fact, we can construct Brownian excursion as the solution to an SDE:

$\mathrm{d}W^+(t) = \left[\frac{1}{W^+(t)} - \frac{W^+(t)}{1-t}\right] \mathrm{d}t + \mathrm{d}B(t),\quad t\in(0,1)$ (**)

for B a standard Brownian motion. I might return in the next post to why this is valid. For now, note that the first drift term pushes the excursion away from zero, while the second term brings it back to zero as $t\rightarrow 1$.

From this, the second drift term is essentially negligible if we care about scaling $W^+(\epsilon)$ as $\epsilon\rightarrow 0$, and we can say that $W^+(\epsilon)=\Theta(\sqrt{\epsilon})$.

So, returning to the random walk, we have

$\frac{S_{\epsilon N}}{\sqrt{\epsilon N}}\stackrel{d}{\approx} \frac{W^+(\epsilon)}{\sqrt{\epsilon}} = \Theta(1).$

At a heuristic level, it’s tempting to try ‘taking $N\rightarrow\infty$ while fixing $\epsilon N$‘, to conclude that there is a well-defined scaling limit for the RW conditioned to stay positive forever. But we came up with this estimate by taking $N\rightarrow\infty$ and then $\epsilon\rightarrow 0$ in that order. So while the heuristic might be convincing, this is not the outline of a valid argument in any way. However, the SDE representation of $W^+$ in the $\epsilon\rightarrow 0$ regime is useful. If we drop the second drift term in (**), we define the three-dimensional Bessel process, which (again, possibly the subject of a new post) is the correct scaling limit we should be aiming for.

Finally, it’s worth observing that the limit $\{T=\infty\}=\lim_{N\rightarrow\infty} \{T>N\}$ is a monotone limit, and so further tools are available. In particular, if we know that the trajectories of the random walk satisfy the FKG property, then we can define this limit directly. It feels intuitively clear that random walks should satisfy the FKG inequality (in the sense that if a RW is large somewhere, it’s more likely to be large somewhere else). You can do a covariance calculation easily, but a standard way to show the FKG inequality applies is by verifying the FKG lattice condition, and unless I’m missing something, this is clear (though a bit annoying to check) when the increments are Gaussian, but not in general. Even so, defining this monotone limit does not tell you that it is non-degenerate (ie almost-surely finite), for which some separate estimates would be required.

A final remark: in a recent post, I talked about the Skorohod embedding, as a way to construct any centered random walk where the increments have finite variance as a stopped Brownian motion. One approach to conditioning a random walk to lie above some discrete function is to condition the corresponding Brownian motion to lie above some continuous extension of that function. This is a slightly stronger conditioning, and so any approach of this kind must quantify how much stronger. In Section 4 of [BL], the authors do this for the random walk associated with the DGFF conditioned to lie above a polylogarithmic curve.

References

[BD] – Bertoin, Doney – 1994 – On conditioning a random walk to stay nonnegative

[BL] – Biskup, Louidor – 2016 – Full extremal process, cluster law and freezing for two-dimensional discrete Gaussian free field

[DIM] – Durrett, Iglehart, Miller – 1977 – Weak convergence to Brownian meander and Brownian excursion

[Ig1] – Iglehart – 1974 – Functional central limit theorems for random walks conditioned to stay positive

[Ig2] – Iglehart – 1974 – Random walks with negative drift conditioned to stay positive

[Ka] – Kaigh – 1976 – An invariance principle for random walk conditioned by a late return to zero

[KS] – Kesten, Stigum – 1966 – A limit theorem for multidimensional Galton-Watson processes

[Sk] – Skorohod – 1955 – Limit theorems for stochastic processes with independent increments

# Skorohod embedding

Background

Suppose we are given a standard Brownian motion $(B_t)$, and a stopping time T. Then, so long as T satisfies one of the regularity conditions under which the Optional Stopping Theorem applies, we know that $\mathbb{E}[B_T]=0$. (See here for a less formal introduction to OST.) Furthermore, since $B_t^2-t$ is a martingale, $\mathbb{E}[B_T^2]=\mathbb{E}[T]$, so if the latter is finite, so is the former.

Now, using the strong Markov property of Brownian motion, we can come up with a sequence of stopping times $0=T_0, T_1, T_2,\ldots$ such that the increments $T_k-T_{k-1}$ are IID with the same distribution as T. Then $0,B_{T_1},B_{T_2},\ldots$ is a centered random walk. By taking T to be the hitting time of $\{-1,+1\}$, it is easy to see that we can embed simple random walk in a Brownian motion using this approach.

Embedding simple random walk in Brownian motion.

The Skorohod embedding question asks: can all centered random walks be constructed in this fashion, by stopping Brownian motion at a sequence of stopping time? With the strong Markov property, it immediately reduces the question of whether all centered finite-variance distributions X can be expressed as $B_T$ for some integrable stopping time T.

The answer to this question is yes, and much of what follows is drawn from, or at least prompted by Obloj’s survey paper which details the problem and rich history of the many approaches to its solution over the past seventy years.

Applications and related things

The relationship between random walks and Brownian motion is a rich one. Donsker’s invariance principle asserts that Brownian motion appears as the scaling limit of a random walk. Indeed, one can construct Brownian motion itself as the limit of a sequence of consistent random walks with normal increments on an increasingly dense set of times. Furthermore, random walks are martingales, and we know that continuous, local martingales can be expressed as a (stochastically) time-changed Brownian motion, from the Dubins-Schwarz theorem.

The Skorohod embedding theorem can be used to prove results about random walks with general distribution by proving the corresponding result for Brownian motion, and checking that the construction of the sequence of stopping times has the right properties to allow the result to be carried back to the original setting. It obviously also gives a coupling between a individual random walk and a Brownian motion which may be useful in some contexts, as well as a coupling between any pair of random walks. This is useful in proving results for random walks which are much easier for special cases of the distribution. For example, when the increments are Gaussian, or when there are combinatorial approaches to a problem about simple random walk. At the moment no aspect of this blog schedule is guaranteed, but I plan to talk about the law of the iterated logarithm shortly, whose proof is approachable in both of these settings, as well as for Brownian motion, and Skorohod embedding provides the route to the general proof.

At the end, we will briefly compare some other ways to couple a random walk and a Brownian motion.

One thing we could do is sample a copy of X independently from the Brownian motion, then declare $T= \tau_{X}:= \inf\{t\ge 0: B_t=X\}$, the hitting time of (random value) X. But recall that unfortunately $\tau_x$ has infinite expectation for all non-zero x, so this doesn’t fit the conditions required to use OST.

Skorohod’s original method is described in Section 3.1 of Obloj’s notes linked above. The method is roughly to pair up positive values taken by X appropriately with negative values taken by X in a clever way. If we have a positive value b and a negative value a, then $\tau_{a,b}$, the first hitting time of $\mathbb{R}\backslash (a,b)$ is integrable. Then we choose one of these positive-negative pairs according to the projection of the distribution of X onto the pairings, and let T be the hitting time of this pair of values. The probability of hitting b conditional on hitting {a,b} is easy to compute (it’s $\frac{-a}{b-a}$) so we need to have chosen our pairs so that the ‘probability’ of hitting b (ie the density) comes out right. In particular, this method has to start from continuous distributions X, and treat atoms in the distribution of X separately.

The case where the distribution X is symmetric (that is $X\stackrel{d}=-X$) is particularly clear, as then the pairs should be $(-x,x)$.

However, it feels like there is enough randomness in Brownian motion already, and subsequent authors showed that indeed it wasn’t necessary to introduce extra randomness to provide a solution.

One might ask whether it’s possible to generate the distribution on the set of pairs (as above) out of the Brownian motion itself, but independently from all the hitting times. It feels like it might be possible to make the distribution on the pairs measurable with respect to

$\mathcal{F}_{0+} = \bigcap\limits_{t>0} \mathcal{F}_t,$

the sigma-algebra of events determined by limiting behaviour as $t\rightarrow 0$ (which is independent of hitting times). But of course, unfortunately $\mathcal{F}_{0+}$ has a zero-one law, so it’s not possible to embed non-trivial distributions there.

Dubins solution

The exemplar for solutions without extra randomness is due to Dubins, shortly after Skorohod’s original argument. The idea is to express the distribution X as the almost sure limit of a martingale. We first use the hitting time of a pair of points to ‘decide’ whether we will end up positive or negative, and then given this information look at the hitting time (after this first time) of two subsequent points to ‘decide’ which of four regions of the real interval we end up in.

I’m going to use different notation to Obloj, corresponding more closely with how I ended up thinking about this method. We let

$a_+:= \mathbb{E}[X \,|\, X>0], \quad a_- := \mathbb{E}[X\,|\, X<0],$ (*)

and take $T_1 = \tau_{\{a_-,a_+\}}$. We need to check that

$\mathbb{P}\left( B_{T_1}=a_+\right) = \mathbb{P}\left(X>0\right),$

for this to have a chance of working. But we know that

$\mathbb{P}\left( B_{T_1}=a_+\right) = \frac{a_+}{a_+-a_-},$

and we can also attack the other side using (*) and the fact that $\mathbb{E}[X]=0$, using the law of total expectation:

$0=\mathbb{E}[X]=\mathbb{E}[X\,|\, X>0] \mathbb{P}(X>0) + \mathbb{E}[X\,|\,X<0]\mathbb{P}(X<0) = a_+ \mathbb{P}(X>0) + a_- \left(1-\mathbb{P}(X>0) \right),$

$\Rightarrow\quad \mathbb{P}(X>0)=\frac{a_+}{a_+-a_-}.$

Now we define

$a_{++}=\mathbb{E}[X \,|\, X>a_+],\quad a_{+-}=\mathbb{E}[X\,|\, 0

and similarly $a_{-+},a_{--}$. So then, conditional on $B_{T_1}=a_+$, we take

$T_2:= \inf_{t\ge T_1}\left\{ B_t\not\in (a_{+-},a_{++}) \right\},$

and similarly conditional on $B_{T_1}=a_-$. By an identical argument to the one we have just deployed, we have $\mathbb{E}\left[B_{T_2} \,|\,\mathcal{F}_{T_1} \right] = B_{T_1}$ almost surely. So, although the $a_{+-+}$ notation now starts to get very unwieldy, it’s clear we can keep going in this way to get a sequence of stopping times $0=T_0,T_1,T_2,\ldots$ where $B_{T_n}$ determines which of the $2^n$ regions of the real line any limit $\lim_{m\rightarrow\infty} B_{T_m}$ should lie in.

A bit of work is required to check that the almost sure limit $T_n\rightarrow T$ is almost surely finite, but once we have this, it is clear that $B_{T_n}\rightarrow B_T$ almost surely, and $B_T$ has the distribution required.

Komlos, Major, Tusnady coupling

We want to know how close we can make this coupling between a centered random walk with variance 1, and a standard Brownian motion. Here, ‘close’ means uniformly close in probability. For large times, the typical difference between one of the stopping times $0,T_1,T_2,\ldots$ in the Skorohod embedding and its expectation (recall $\mathbb{E}[T_k]=k$) is $\sqrt{n}$. So, constructing the random walk $S_0,S_1,S_2,\ldots$ from the Brownian motion via Skorohod embedding leads to

$\left |S_k - B_k \right| = \omega(n^{1/4}),$

for most values of $k\le n$. Strassen (1966) shows that the true scale of the maximum

$\max_{k\le n} \left| S_k - B_k \right|$

is slightly larger than this, with some extra powers of $\log n$ and $\log\log n$ as one would expect.

The Komlos-Major-Tusnady coupling is a way to do a lot better than this, in the setting where the distribution of the increments has a finite MGF near 0. Then, there exists a coupling of the random walk and the Brownian motion such that

$\max_{k\le n}\left|S_k- B_k\right| = O(\log n).$

That is, there exists C such that

$\left[\max_{k\le n} \left |S_k-B_k\right| - C\log n\right] \vee 0$

is a tight family of distributions, indeed with uniform exponential tail. To avoid digressing infinitely far from my original plan to discuss the proof of the law of iterated logarithm for general distributions, I’ll stop here. I found it hard to find much coverage of the KMT result apart from the challenging original paper, and many versions expressed in the language of empirical processes, which are similar to random walks in many ways relevant to convergence and this coupling, but not for Skorohod embedding. So, here is a link to some slides from a talk by Chatterjee which I found helpful in getting a sense of the history, and some of the modern approaches to this type of normal approximation problem.

# DGFF 2 – Boundary conditions and Gibbs-Markov property

In the previous post, we defined the Discrete Gaussian Free Field, and offered some motivation via the discrete random walk bridge. In particular, when the increments of the random walk are chosen to be Gaussian, many natural calculations are straightforward, since Gaussian processes are well-behaved under conditioning and under linear transformations.

Non-zero boundary conditions

In the definition of the DGFF given last time, we demanded that $h\equiv 0$ on $\partial D$. But the model is perfectly well-defined under more general boundary conditions.

It’s helpful to recall again the situation with random walk and Brownian bridge. If we want a Brownian motion which passes through (0,0) and (1,s), we could repeat one construction for Brownian bridge, by taking a standard Brownian motion and conditioning (modulo probability zero technicalities) on passing through level s at time 1. But alternatively, we could set

$B^{\mathrm{drift-br}}(t) = B(t)+ t(s-B(1)),\quad t\in[0,1],$

or equivalently

$B^{\mathrm{drift-br}}(t)=B^{\mathrm{br}}(t)+ st, \quad t\in[0,1].$

That is, a Brownian bridge with drift can be obtain from a centered Brownian bridge by a linear transformation, and so certainly remains a Gaussian process. And exactly the same holds for a discrete Gaussian bridge: if we want non-zero values at the endpoints, we can obtain this distribution by taking the standard centred bridge and applying a linear transformation.

We can see how this works directly at the level of density functions. If we take $0=Z_0,Z_1,\ldots,Z_{N-1},Z_N=0$ a centred Gaussian bridge, then the density of $Z=\mathbf{z}\in \mathbb{R}^{N+1}$ is proportional to

$\mathbf{1}\{z_0=z_N=0\}\exp\left( -\frac12 \sum_{i=1}^N (z_i-z_{i-1})^2 \right).$ (3)

So rewriting $z_i= y_i- ki$ (where we might want $k=s/N$ to fit the previous example), the sum within the exponent rearranges as

$-\frac12 \sum_{i=1}^N (y_i-y_{i-1} - k)^2 = -\frac12 \sum_{i=1}^N (y_i-y_{i-1})^2 - 2k(y_N-y_0)+ Nk^2.$

So when the values at the endpoints $z_0,z_n,y_0,y_N$ are fixed, this middle term is a constant, as is the final term, and thus the density of the linearly transformed bridge has exactly the same form as the original one.

In two or more dimensions, the analogue of adding a linear function is to add a harmonic function. First, some notation. Let $\varphi$ be any function on $\partial D$. Then there is a unique harmonic extension of $\varphi$, for which $\nabla \varphi=0$ everywhere on D, the interior of the domain. Recall that $\nabla$ is the discrete graph Laplacian defined up to a constant by

$(\nabla \varphi) _x = \sum\limits_{x\sim y} \varphi_x - \varphi_y.$

If we want $h^D$ instead to have boundary values $\varphi$, it’s enough to replace $h^D$ with $h^D+\varphi$. Then, in the density for the DGFF ( (1) in the previous post), the term in the exponential becomes (ignoring the $\frac{1}{4d}$ )

$-\sum\limits_{x\sim y} \left[ (h^D_x-h^D_y)^2 + (\varphi_x-\varphi_y)^2 +2(h^D_x - h^D_y)(\varphi_x-\varphi_y)\right].$

For each $x\in D$, on taking this sum over its neighbours $y\in \bar D$, the final term vanishes (since $\varphi$ is harmonic), while the second term is just a constant. So the density of the transformed field, which we’ll call $h^{D,\varphi}$ is proportional to (after removing the constant arising from the second term above)

$\mathbf{1}\left\{h^{D,\varphi}_x = \varphi_x,\, x\in\partial D\right\} \exp\left( -\frac{1}{4d} \sum\limits_{x\sim y} \left( h^{D,\varphi}_x - h^{D,\varphi}_y \right)^2 \right).$

So $h^{D,\varphi}:= h^D + \varphi$ satisfies the conditions for the DGFF on D with non-zero boundary conditions $\varphi$.

Harmonic functions and RW – a quick review

Like the covariances in DGFF, harmonic functions on D are related to simple random walk on D stopped on $\partial D$. (I’m not claiming a direct connection right now.) We can define the harmonic extension $\varphi$ to an interior point x by taking $\mathbb{P}_x$ to be the law of SRW $x=Z_0,Z_1,Z_2,\ldots$ started from x, and then setting

$\varphi(x):= \mathbb{E}\left[ \varphi_{\tau_{\partial d}} \right],$

where $\tau_{\partial D}$ is the first time that the random walk hits the boundary.

Inverse temperature – a quick remark

In the original definition of the density of the DGFF, there is the option to add a constant $\beta>0$ within the exponential term so the density is proportional to

$\exp\left(-\beta \sum\limits_{x\sim y} (h_x-h_y)^2 \right).$

With zero boundary conditions, the effect of this is straightforward, as varying $\beta$ just rescales the values taken by the field. But with non-zero boundary conditions, the effect is instead to vary the magnitude of the fluctuations of the values of the field around the (unique) harmonic function on the domain with those BCs. In particular, when $\beta\rightarrow \infty$, the field is ‘reluctant to be far from harmonic’, and so $h^D \Rightarrow \varphi$.

This parameter $\beta$ is called inverse temperature. So low temperature corresponds to high $\beta$, and high stability, which fits some physical intuition.

A Markov property

For a discrete (Gaussian) random walk, the Markov property says that conditional on a given value at a given time, the trajectory of the process before this time is independent of the trajectory afterwards. The discrete Gaussian bridge is similar. Suppose we have as before $0=Z_0,Z_1,\ldots, Z_N=0$ a centred Gaussian bridge, and condition that $Z_k=y$, for $k\in\{1,\ldots,N-1\}$, and $y\in\mathbb{R}$. With this conditioning, the density (3) splits as a product

$\mathbf{1}\{z_0=z_N=0, z_k=y\}\exp\left(-\frac12 \sum\limits_{i=1}^N (z_i-z_{i-1})^2 \right) =$

$\mathbf{1}\{z_0=0,z_k=y\} \exp\left(-\frac12 \sum\limits_{i=1}^k (z_i-z_{i-1})^2 \right) \cdot \mathbf{1}\{z_k=y,z_N=0\} \exp\left(-\frac12 \sum\limits_{i=k+1}^N (z_i-z_{i-1})^2 \right).$

Therefore, with this conditioning, the discrete Gaussian bridge splits into a pair of independent discrete Gaussian bridges with drift. (The same would hold if the original process had drift too.)

The situation for the DGFF is similar, though rather than focusing on the condition, it makes sense to start by focusing on the sub-domain of interest. Let $A\subset D$, and take $B=\bar D\backslash A$. So in particular $\partial A\subset B$.

Then we have that conditional on $h^D\big|_{\partial A}$, the restricted fields $h^D\big|_{B\backslash \partial A}$ and $h^D\big|_A$ are independent. Furthermore, $h^D\big|_A$ has the distribution of the DGFF on A, with boundary condition given by $h^D\big|_{\partial A}$. As in the discrete bridge, this follows just by splitting the density. Every gradient term corresponds to an edge in the underlying graph that lies either entirely inside $\bar A$ or entirely inside B. This holds for a general class of Gibbs models where the Hamiltonian depends only on the sum of some function of the heights (taken to be constant in this ‘free’ model) and the sum of some function of their nearest-neighbour gradients.

One additional and useful interpretation is that if we only care about the field on the restricted region A, the dependence of $h^D\big|_A$ on $h^D\big|_{D\backslash A}$ comes only through $h^D\big|_{\partial A}$. But more than that, it comes only through the (random) harmonic function which extends the (random) values taken on the boundary of A to the whole of A. So, if $h^A$ is an independent DGFF on A with zero boundary conditions, we can construct the DGFF $h^D$ from its value on $D\backslash A$ via

$h^D_x \stackrel{d}= h^A_x + \varphi^{h^D\big|_{\partial A}},$

where $\varphi^{h^D\big|_{\partial A}}$ is the unique harmonic extension of the (random) values taken by $h^D$ on $\partial A$ to $\bar A$.

This Markov property is crucial to much of the analysis to come. There are several choices of the restricted domain which come up repeatedly. In the next post we’ll look at how much one can deduce by taking A to be the even vertices in D (recalling that every integer lattice $\mathbb{Z}^d$ is bipartite), and then taking A to be a finer sublattice within D. We’ll use this to get some good bounds on the probability that the DGFF is positive on the whole of D. Perhaps later we’ll look at a ring decomposition of $\mathbb{Z}^d$ consisting of annuli spreading out from a fixed origin. Then the distribution of the field at this origin can be considered, via the final idea discussed above, as the limit of an infinite sequence of random harmonic functions given by the values taken by the field at increasingly large radius from the origin. Defining the DGFF on the whole lattice depends on the existence or otherwise of this local limit.

# DGFF 1 – The discrete Gaussian free field from scratch

I’ve moved to Haifa in northern Israel to start a post-doc in the probability group at the Technion, and now that my thesis is finished I want to start blogging again. The past couple of weeks have been occupied with finding an apartment and learning about the Discrete Gaussian Free Field. All questions about the apartment are solved, but fortunately lots remain open about the DGFF, so I thought I’d write some background about this object and methods which have been used to study it.

Background – Random walk bridge

When we think of a random walk, we usually think of the index as time, normally going forwards. So for a random walk bridge, we might assume $Z_0=0$, and then condition on $Z_N=0$, thinking of this as a demand that the process has returned to zero at the future time. In some applications, this is the ideal intuition, but in others, it is more useful to think of the random walk bridge

$(0=Z_0,Z_1,\ldots,Z_{N-1},Z_N=0),$

as a random height function indexed by [0,N], where the probability of a given path decomposes naturally into a product depending on the N increments, up to a normalising constant.

Naturally, we are interested in the asymptotic behaviour of such a random walk bridge when $N\rightarrow\infty$. So long as the step distribution has finite variance, a conditioned version of Donsker’s theorem shows that the rescaled random walk bridge converges in distribution to Brownian bridge. Note that Brownian bridge

$(B^{\mathrm{br}}_t, t\in[0,1])$

can be constructed either by conditioning a standard Brownian motion B to return to zero at time one (modulo some technicalities – this event has zero probability), or by applying an appropriate (random) linear shift

$B^{\mathrm{br}}(t):= B(t) - tB(1).$ (*)

It is not too hard to calculate the distribution of $B^{\mathrm{br}}(t)$ for each $t\in[0,1]$, and with a bit more work, one can calculate the joint distribution of $(B^{\mathrm{br}}(s),B^{\mathrm{br}}(t))$. In particular, the joint distribution is multivariate Gaussian, and so everything depends on the covariance ‘matrix’ (which here is indexed by [0,1]).

So if we return to a random walk bridge what should the step distribution be? Simple symmetric RW is a natural choice, as then lots of the quantities we might want to consider boil down to combinatorial calculations. Cleverness and Stirling’s formula can often get us useful asymptotics. But there are lots of inconveniences, not least the requirement to be careful about parity (N has to be even for a start unless you make the walk lazy, in which case the combinatorics becomes harder), and even if these can be overcome in a given calculation, it would be better not to have this.

The claim is that the random walk with Gaussian increments is by far the easiest to analyse asymptotically. As a further heuristic, think about the statement of the central limit theorem in the case where the underlying distribution is normal: it’s true but obvious. [Indeed, it’s my favourite piece of advice to anyone taking second year probability exams to check that your proposed statement of CLT does actually work for $N(\mu,\sigma^2)$…] More concretely, if a RW has Gaussian increments, then the path $(Z_1,\ldots,Z_N)$ is a multivariate normal, or a Gaussian process with finite index set. In particular, covariances define the distribution. It remains a Gaussian process after conditioning on $Z_N=0$, and the linear tilting argument at (*) remains true here, and can indeed be applied to turn any boundary conditions into any other boundary conditions.

The discrete Gaussian free field

We know how to generalise the domain of a random walk to higher dimensions. But what generalising the index to higher dimension? So now there is definitely no arrow of time, and the notion of a random height function above $\mathbb{Z}^2$ (or a subset of it) is helpful, for which a scaling limit might be a random surface rather than Brownian motion.

Because we can’t well-order $\mathbb{Z}^d$, it’s harder to define any such random object on the entire lattice immediately, so we start with compact connected subsets, with zero boundary conditions, as in the one-dimensional case of random walk bridge. Formally, let D be a finite subset of $\mathbb{Z}^d$, and the boundary $\partial D$ those elements of $D^c$ which are adjacent to an element of D, and let $\bar D:= D\cup \partial D$.

Then, the discrete Gaussian free field on D is a random real vector $h^D=(h^D_x: x\in \bar D)$, with probability density proportional to

$\mathbf{1}\{h^D_x=0, x\in\partial D\}\exp\left ( - \frac{1}{4d} \sum_{x\sim y}(h^D_x - h^D_y)^2 \right),$ (1)

where we write $x\sim y$ if that x,y are adjacent in $\bar D$. We won’t at any stage worry much about the partition function which normalises this pdf. Note also that $\frac{1}{4d}$ is just a convenient choice of constant, which corresponds to one of the canonical choices for the discrete Laplacian. Adjusting this constant is the same as uniformly rescaling the values taken by the field.

The immediate interpretation of (1) is that the values taken by the field at vertices which are close to each other are positively correlated. Furthermore, the form of the density is Gaussian. Concretely, if the values of $h^D$ are fixed everywhere except one vertex $x\in D$, then the conditional distribution of $h^D_x$ is Gaussian. Later, or in subsequent posts, we will heavily develop this idea. Alternatively, we could if we really wanted describe the model in terms of independent Gaussians describing the ‘increment’ along each edge in D (which we should direct), subject to a very large number of conditions, namely that the sum of increments along any directed cycle is zero. This latter description might be more useful if you wanted to define a DGFF on a more sparse graph, but won’t be useful in what follows.

Note that we can rearrange the Laplacian in (1) in terms of the transition kernel p( ) of the simple random walk of D to obtain

$\exp\left( -\frac12 (h^D)^T (\mathbf{P}-\mathbf{1})h^D \right),$

where $P_{x,y}=p(y-x)$ is the transition matrix of SRW on D. In particular, this means that the free field is Gaussian, and we can extract the covariances via

$\mathrm{Cov}(h^D_x,h^D_y) = \left[ (\mathbf{1}-\mathbf{P})^{-1}\right]_{x,y}$

$= \left[\sum_{n\ge 0} \mathbf{P}^n\right]_{x,y} = \sum_{n\ge 0} \mathbb{P}_x\left[X_n=y,\tau_{\partial D}>n\right],$

where, under $\mathbb{P}_x$, $(X_0,X_1,\ldots)$ is simple random walk started from x.

This final quantity records the expected number of visits to y before leaving the domain D, for a random walk started at x, and is called the Green’s function.

In summary, the DGFF on D is the centred Gaussian random vector indexed by $\bar D$ with covariance given by the Green’s function $G_D(x,y)$.

How many of these equivalences carries over to more general D-indexed random fields is discussed in the survey paper by Velenik. But it’s worth emphasising that having the covariance given by the Green’s function as in the definition we’ve just given is a very nice property, as there are lots of pre-existing tools for calculating these. By contrast, it’s hard to think of a natural model for an integer-valued surface of this kind, as an analogue to SRW.

[Though definitely not impossible. The nicest example I’ve heard of is for height functions of large uniform domino tilings within their ‘arctic circle’, which have GFF asymptotics. See this paper by Kenyon.]

A continuous limit?

We motivated the discussion of random walk bridge by the limit object, namely Brownian bridge. Part of the reason why the DGFF is more interesting than Gaussian random walk bridge, is that the limit object, the (continuum) Gaussian free field is hard to define classically in two dimensions.

We might suppose that the DGFF in $V_N$, the square box of width N has some scaling limit as $N\rightarrow\infty$. However, for fixed $x,y\in [0,1]^2$, (and taking integer parts component-wise), well-known asymptotics for SRW in a large square lattice (more on this soon hopefully) assert that

$\mathrm{Cov}(h^{V_N}_{\lfloor Nx \rfloor},h^{V_N}_{\lfloor Ny\rfloor}) \sim \log |x-y|,$ (2)

and so any scaling limit will rescale only the square domain, not the height (since there is no N on the RHS of (2)). However, then the variance of the proposed limit is infinite everywhere.

So the GFF does not exist as a random height function on $[0,1]^2$, with the consequence that a) more care is needed over its abstract definition; b) the DGFF in 2D on a large square is an interesting object, since it does exist in this sense.

What makes it ‘free’?

This seemed like a natural question to ask, but I’ve received various answers. Some sources seem to suggest that having zero boundary condition is free. Other sources refer to the Hamiltonian (that is the term inside the exponential function at (1) ) as free since it depends only on the increments between values. If the Hamiltonian also depends on the heights themselves, for example via the addition of a $\sum_{x} \Psi(h^D_x)$ term, then for suitable choice of function $\Psi$, this is interpreted as a model where the particles have mass. The physical interpretation of these more general Gibbs measures is discussed widely, and I’m not very comfortable with it all at the moment, but aim to come back to it later, when hopefully I will be more comfortable.

# Parking on a ring, linear hashing

I’ve spent most of my doctorate trying to analyse how adding destructive dynamics affects the behaviour of a particular random growth process, the classical random graph. In this post I’m going to talk about another random growth process, which is slightly less natural, but for which one can show some similar qualitative properties.

The model, and the additive coalescent

Consider m places arranged in a circle, and for consistency of analogy we think of these as parking spaces. Some number n of cars will arrive one at a time. Each car will arrive at a space chosen uniformly at random. If it is empty they will park in it, otherwise they will look clockwise until they find an empty space, and park there. For now we are only interested in growth, so we assume cars never leave. We are interested in the sizes of blocks of consecutively parked cars.

The reason to consider this slightly unnatural statement is its equivalence to the problem of hashing with linear probing, apparently a key topic in computer science, which I won’t pretend that I know anything about. In any case, it’s a nice model, and it seems reasonable that it would have a basis in more realistic search algorithms.

So, how does the sequence of sizes of blocks of consecutively parked cars grow? Well, given the sequence of block sizes, it is reasonably easy to convince yourself that the order of the blocks around the circle is uniformly random, and the number of empty spaces between adjacent blocks is also uniformly random.

Assume for now that there are at least three blocks. A block of size x can merge with a block of size y with the arrival of the next car only if the blocks are adjacent, with exactly one empty space between them. The chance of this is uniform among all pairs of blocks. Now suppose this is the case, and that the block of size y lies clockwise from the block of size x. Then they will merge precisely if the next car arrives at any of the x occupied spaces in that block, or at the empty space between the pair of blocks. This has probability $\frac{x+1}{m}$. There’s also the opposite ordering to consider, where the block of size x lies clockwise from the other. The total probability of this merge $\{x,y\}\mapsto \{x+y+1\}$ is therefore proportional to (x+y+2).

So the process of block sizes looks a bit like the additive coalescent, at least for large blocks. This is in contrast to the random graph process, where the sequence of component sizes behaves exactly like a multiplicative coalescent, where blocks merge at a rate proportional to the product of their sizes.

Asymptotics

As in the random graph process, it’s interesting to ask roughly how large the largest block will be in such a configuration. Pittel [3] considers the case where the number of empty places $\ell = m-n \approx \beta m$, for some $\beta\in (0,1)$.

A less interesting model would be to choose the positions of the n cars uniformly at random. But then the size of a block is roughly geometric with parameter $\beta$, and there are $\Theta(m)$ blocks with high probability. Relatively straightforward calculations in extreme value theory suggest that the largest block is likely to have size on the order of $\log m$ in this setting.

Of course, the actual model is slightly more complicated, because the size of a block is self-reinforcing, since larger blocks are more likely to grow than smaller blocks. However, we can still get somewhere with naïve estimates. Let’s label the places clockwise. Then in order for there to be a block starting at 0 and stretching beyond $\alpha \log m$, a necessary condition is that at least $\alpha \log m$ cars arrive at those places. The number of cars which arrive at those places is binomial, since there are n cars, and each arrives at a place chosen uniformly, and independently of the other cars. So this event corresponds to

$\mathrm{Bin}(n,\frac{\alpha \log m}{m}) \ge \alpha \log m.$

Then, since $n\approx (1-\beta)n$, this event corresponds approximately to

$\mathrm{Po}((1-\beta)\alpha \log m) \ge \alpha \log m.$

The probability that a Poisson RV is at least a constant multiple larger than its mean decays exponentially with the mean, hence in this case the probability is asymptotically some negative power of m, depending on the value of $\alpha$. But there are $O(m)$ possible places for such a block to start, so whether we can apply a union bound usefully or not depends on whether the power of m is strictly less than -1.

Since all of this depends on $\alpha$, it is reasonable that everything is fine, and the largest block does have size at least $\alpha \log m$ when $\alpha$ is small, and very unlikely when $\alpha$ is large. This heuristic argument fits with Pittel’s theorem. Indeed, his result shows much stronger concentration: that the fluctuations of the size of the largest block are O(1).

Critical regime and empirical processes

The following is a paraphrase of the introduction and some methods from [2].

Obviously, once m=m cars have arrived, there’s no room for manoeuvre and definitely all the places are taken in one giant block. But it’s not obvious in general what scaling for the number of gaps will give rise to giant blocks of $\Theta(m)$ cars.

As for the random graph, we can find a process similar to the exploration process of a (random) graph which encodes much of the information we care about. Let $Y_k$ be the number of cars which arrive at place k. So the sum of the $Y_k$s will be n, the total number of cars. Now consider the process

$C_0=0, \ldots, C_{k+1}=C_k + Y_{k+1}-1.$

A block has the property that the number of arrivals within that set of places is equal to the number of places. So every time this *empirical process* C drops below its previous running minimum, this indicates the end of a block. To make this equivalence precise, we need to be a bit careful about where we start counting. It works exactly if we start at the beginning of a block. If not, it might introduce some unwanted divisions within the first block.

What we have is a process that looks roughly like a random walk that is constrained to pass through the point (m,n-m), which is equal to (m,-l). Even if we aren’t totally precise about how this is like a random walk, we would expect to see Brownian fluctuations after rescaling. Indeed, we might expect to see a Brownian bridge added to a deterministic linear function with negative gradient. But this is only meaningful if the random part is at least as large as the deterministic part, and since the fluctuations have order $\sqrt{m}$, if l is much larger than this, the rescaled empirical process is essentially deterministic, so we won’t see any macroscopic excursions above the minimum.

If l is substantially smaller than $\sqrt{m}$, then there is no real difference between (m,-l) and (m,0), and what we see is just a Brownian bridge. At this point, where we choose to start the process is actually important. If we were to start it at the minimum of the Brownian bridge instead, we would have seen a Brownian excursion, which corresponds to one block occupying (almost) all of the places.

Unsurprisingly, the story is completed by considering $\ell=\Theta(\sqrt{m})$, where the rescaled empirical process looks like a slanted Brownian bridge, that is Brownian motion conditioned to pass through $(1,-\frac{\ell}{\sqrt{m})$. There isn’t an obvious fix to the question of where to start the process, but it turns out that the correct way is now adding a Brownian excursion onto the deterministic linear function with gradient $- \frac{\ell}{\sqrt{m}}$. It’s now reasonable that the excursions above the minimum should macroscopic.

This scaling limit works dynamically as well, where the same Brownian excursion is used for different gradients of the deterministic line, corresponding to $\ell$ moving through the critical window $m-\Theta(\sqrt{m})$. Finally, a direction to Bertoin’s recent paper [1] for the model with an additional destructive property. Analogous to the forest fire, blocks of cars are removed at a rate proportional to their size (as a result, naturally, of ‘Molotov cocktails’…). Similar effects of self-organised criticality are seen when the rate of bombs is scaled appropriately.

References

[1] – Bertoin – Burning cars in a parking lot (paper / slides)

[2] – Chassaing + Louchard – Phase transition for parking blocks, Brownian excursion and coalescence (arXiv)

[3] – Pittel – Linear probing: the probable largest search time grows logarithmically with the number of records

# Fair games and the martingale strategy III

Gambler’s Ruin

Continuing directly from the previous post, the nicest example of the optional stopping theorem we developed there is to example a simple random walk constrained between two values, say 0 and N. This represents an idealised gambling situation, where the gambler stops playing either when they reach some pre-agreed profit, or when they go bankrupt. We assume that we start at level k, for k = 1,2,…,N-1.

Naturally, we want to know the probabilities of winning (ie getting to N) and losing (ie going bankrupt). We could set this up by conditioning on the first step. Let $p_k$ be the probability of winning starting from level k. Then we must have

$p_k= \frac12 p_{k+1}+\frac12 p_{k-1},\quad k=1,\ldots,N-1,$ (*)

with the obvious boundary conditions $p_0=0, p_N=1$. In an ideal world, we just know how to solve second order difference equations like (*). Well, actually it isn’t too hard, because we can see from (*) directly that

$p_{k+1}-p_k = p_k-p_{k-1},$

and so $p_k$ is a linear function of k, and so $p_k = k/N$ follows pretty much immediately.

But, we can also use OST profitably. Let T be the time at which we first hit 0 or N. It’s intuitively clear that this should have finite expectation, since the problems you might encounter with just the hitting time of a single level shouldn’t apply. Or you can consider the expected number of steps before you see N ups or downs in a row, which certainly provides an upper bound on T. This random number of steps is sort of geometric (at least, can be upper bounded by a geometric RV) and so has finite expectation. So can apply OST to X at T, and we have

$\mathbb{E}[X_T] = N\cdot \mathbb{P}(X_T=N) + 0 \cdot \mathbb{P}(X_T=0) = \mathbb{E}[X_0]=k,$

from which we also derive $p_k=k/N$.

The reason we talk about gambler’s ruin is by considering the limit $N\rightarrow\infty$ with k fixed. After a moment’s thought, it’s clear we can’t really talk about stopping the process when we hit infinity, since that won’t happen at any finite time. But we can ask what’s the probability that we eventually hit zero. Then, if we imagine a barrier at level N, the probability that we hit 0 at some point is bounded below by the probability that we hit 0 before we hit level N (given that we know we hit either zero or level N with probability one), and this is $\frac{N-k}{N}$, and by choosing N large enough, we can make this as close to 1 as we want. So the only consistent option is that the probability of hitting 0 at some point is one. Hence gambler’s ruin. With probability one, ruin will occur. There’s probably a moral lesson hiding there not especially subtly.

A problem about pricing options

So the deal here seems to be that if you just care about your average, it doesn’t matter how to choose to play a sequence of fair games. But what if you care about something other than your average? In any real setting, we maybe care about slightly more than this. Suppose I offer you a bet on a coin toss: you get £3 if it comes up heads, and I get £1 if it comes up tails. Sounds like a good bet, since on average you gain a pound. But what about if you get £10,003 if it comes up heads and I get £10,001 if it comes up tails? I’m guessing you’re probably not quite so keen now.

But if you were an international bank, you might have fewer reservations about the second option. My intention is not to discuss whether our valuation of money is linear here, but merely to offer motivation for the financial option I’m about to propose. The point is that we are generally risk-averse (well, most of us, most of the time) and so we are scared of possible large losses, even when there is the possibility of large profits to balance it out.

Let’s assume we have our simple random walk, and for definiteness let’s say it starts at £1. Suppose (eg as a very niche birthday present) we have the following opportunity: at any point between now and time t=5, we have the right to buy one unit of the stock for £2.

We want to work out how much this opportunity, which from now on I’m going to call an option, is worth on average. Note that now it does seem that when we choose to cash in the option will have an effect on our return, and so we will have to include this in the analysis.

Note that, once we’ve bought a unit of the stock, we have an asset which is following a simple random walk (ie sequential fair games) and so from this point on its expected value remains unchanged. So in terms of expectation, we might as well sell the stock at the same moment we buy it. So if we cash in the option when the stock is currently worth £X, we will on average have a return of £(X-2). This means that we’ll only ever consider exercising our option if the current value of the stock is greater than £2. This narrows down our strategy slightly.

This sort of option minimises the risk of a large loss, since the worst thing that happens is that you never choose to exercise your option. So if you actually paid for the right to have this option, that cost is the largest amount you can lose. In the trading world, this type of opportunity is called an American option.

The trick here is to work backwards in time, thinking about strategies. If at time t=4, the stock is worth £1, then the best that can happen is that it’s worth £2 at time t=5, and this still gains you no wealth overall. Similarly if it’s worth £0 at time t=3. So we’ve identified a region where, if the stock value enters this region, we might as well rip up our contract, because we definitely aren’t going to gain anything. Remember now that we’ve also said you won’t ever cash in if the stock’s value is at most £2, because you don’t gain anything on average.

Now suppose that the stock has value £3 at time t=4. There’s no danger of it ever getting back below £2 during the lifetime of the option, so from now on your potential return is following the trajectory of a simple random walk, ie a fair game. So on average, it makes no difference whether you cash in now, or wait until t=5, or some combination of the two. The same argument holds if the stock has value £4 at time t=3 or time t=4, and so we can identify a region where you might as well cash in.

What about the final region? If the stock value is greater than £2, but not yet in the definitely-cash-in area, what should you do? Well, if you think about it, the value of the stock is a fair game. But your return should be better than that, because the stock price doesn’t take account of the fact that you wouldn’t buy in (and make a loss overall) if the value drops below £2. So at this stage, your future options are better than playing a fair game, and so it doesn’t make sense (in terms of maximising your *average*) to cash in.

Now we can actually work backwards in time to establish how much any starting value is worth under this optimal strategy. We can fill in the values in the ‘doomed’ area (ie all zeros) and on the ‘cash in now’ area (ie current value minus 2), and construct backwards using the fact that we have a random walk.

The final answer ends up being 7/16 if the stock had value £1 at time 0. Note that the main point here is that working out the qualitative form of the strategy was the non-trivial part. Once we’d done that, everything was fairly straightforward. I claim that this was a reasonably fun adjustment to the original problem, but have minimal idea whether pricing options is in general an interesting thing to do.

Anyway, I hope that provided an interesting overview to some of the topics of interest within the question of how to choose strategies for games based on random processes.

# Fair games and the martingale strategy I

I went back to my school a couple of weeks ago and gave a talk. I felt I’d given various incarnations of a talk on card-shuffling too many times, so it was time for a new topic. The following post (and time allowing, one or two more) is pretty much what I said.

The Martingale Strategy

Suppose we bet repeatedly on the outcome of tossing a fair coin. Since it’s November, my heart is set on buying an ice cream that costs £1, so my aim is to win this amount from our game. My strategy is this:

First, I bet £1. If I win, then that’s great, because I now have made exactly enough profit to buy the ice cream. If I lose, then I play again, and this time I bet £2. Again, if I win, then my total profit is £2-£1 = £1, so I stop playing and buy the ice cream. If I lose, then I play a third time, again doubling my stake. So if I win for the first time on the seventh go, my overall profit will be

£64 – (£1+£2+£4+£8+£16+£32) = £1,

and it’s clear that this can be continued and I will eventually win a round, and at this point my total profit will be £1. So I will always eventually be able to buy my ice cream.

But, there’s nothing special about the value £1, so I could replace the words ‘ice cream’ with ‘private tropical island’, so why am I still here in the UK on a wet Monday when I could be on my beach lounger?

There are some fairly obvious reasons why the strategy I’ve described is not actually a fail-safe way to make a profit. For a start, although with probability one a head will come up eventually, there is a small positive chance that the first 200 rolls will all be tails. At this point, I would have accrued a debt of roughly $2^{200}$ pounds, and this is slightly more than the number of atoms in the universe. All this for an ice cream?

So there are major problems carrying out this strategy in a finite world. And of course, it’s no good if we stop after a very large but finite number of turns, because then there’s always this very small chance that we’ve made a very large loss, which is bad, partly because we can’t have the ice cream, but also because it exactly cancels out the chance of making our £1 profit, and so our overall average profit is exactly zero.

Though I’ve set this up in an intentionally glib fashion, as so often is the case, we might have stumbled across an interesting mathematical idea. That is, if we play a fair game a finite number of times, we have a fair game overall, meaning our overall average profit is zero. But if we are allowed to play a potentially infinite number of times, then it’s not clear how to define our overall ‘average’ profit, since we feel it ought to be zero, as an extension of the finite case, but also might be positive, because it ends up being £1 with probability one.

It’s tempting at this stage to start writing statements like

$1 \times 1 + (-\infty) \times 0=0 ,$

to justify why this might have come about, where we consider the infinitely unlikely event that is infinitely costly. But this is only convincing at the most superficial level, and so it makes more sense to think a bit more carefully about under exactly what circumstances we can extend our observation about the overall fairness of a finite sequence of individual fair games.

A second example

The previous example was based upon a series of coin tosses, and we can use exactly the same source of randomness to produce a simple random walk. This is a process that goes up or down by 1 in each time step, where each option happens with probability ½, independently of the history.

We could avoid the requirement to deal with very large bets by always staking £1, and then cashing in the first time we have a profit of £1. Then, if we start the random walk at zero, it models our profit, and we stop the first time it gets to 1. It’s not obvious whether we hit 1 with probability one. Let’s show this.

In order to hit some positive value k, the random walk must pass through 1, 2, and so on, up to (k-1) and then finally k. So $\mathbb{P}(\text{hit k}) = [\mathbb{P}(\text{hit 1})]^k$. And similarly for negative values. Also, the probability that we return to zero is the same as the probability that we ever hit 1, since after one time-step they are literally the same problem (after symmetry). So, if the probability of hitting 1 is p<1, then the number of visits to zero is geometric (supported on 1,2,3,…) with parameter p, and so

$\mathbb{E}[\text{visits to k}] = \mathbb{E}[\text{visits to zero}] \times \mathbb{P}(\text{hit k})=(1+1/p) \times p^{|k|} = (p+1)p^{|k|-1}.$

Thus, when we sum over all values of k, we are summing a pair of geometric series with exponent <1, and so we get a finite answer. But if the expected number of visits to anywhere (ie the sum across all places) is finite, this is clearly ridiculous, since we are running the process for an infinite time, and at each time-step we must be somewhere! So we must in fact have p=1, and thus another potential counter-example to the claim that a sequence of fair games can sometimes be unfair.

We might have exactly the same set of practical objections, such as this method requiring arbitrarily large liquidity (even though it doesn’t grow exponentially fast so doesn’t seem so bad).

What will actually turn out to be useful is that although the bets are now small, the average time until we hit 1 is actually infinite. Remember that, even though most things we see in real life don’t have this property, it is completely possible for a random variable to take finite values yet have infinite expectation.

Notes on the Martingale Strategy

There’s no reason why the originally proposed strategy had to be based upon fair coin tosses. This strategy might work in a more general setting, where the chance of winning on a given turn is not ½, or is not even constant. So long as at each stage you bet exactly enough that, if you win, you recoup all your losses so far, and one extra pound, this has the same overall effect.

Of course, we need to check that we do eventually win a round, which is not guaranteed if the probability of winning (conditional on not having yet won) decays sufficiently fast. If we let $p_k$ be the probability of winning on turn k, given that we haven’t previously won, then we require that the probability of never winning $\prod_{k\ge 1}(1-p_k)=0$. By taking logs and taking care of the approximations, it can be seen that the divergence or otherwise of $\sum p_k$ determines which way this falls.

In the next post, we’ll talk about how the two problems encountered here, namely allowing large increments, and considering a stopping time with infinite expectation are exactly the two cases where something can go wrong. We’ll also talk about a slightly different setting, where the choice of when to stop playing becomes a bit more dynamic and complicated.